Use symmetry in double integral.

[PsychonautQQ]

Homework Statement

Evaluate the double integral of (2+xy^2) over dA (dxdy) using symmetry where R = [0,1] x [-1,1]

Homework Equations

The Attempt at a Solution

I don't know how to use symmetry to evaluate this.. However if I integrate this integral normally
i first get [2x+2yx^3/3] between 0 and 1. Then i'd take the integral of that between -1 and 1? Where does symmetry come into play? Thanks :D

 

[Mark44]

Homework Statement

Evaluate the double integral of (2+xy^2) over dA (dxdy) using symmetry where R = [0,1] x [-1,1]

Homework Equations

The Attempt at a Solution

I don't know how to use symmetry to evaluate this.. However if I integrate this integral normally
i first get [2x+2yx^3/3] between 0 and 1. Then i'd take the integral of that between -1 and 1? Where does symmetry come into play? Thanks :D

If you let f(x, y) = 2 + xy², then f(x, -y) = f(x, y). This means that the function is symmetric across the x-z plane. However the graph appears over the upper half of region R is the mirror image of how it appears over the lower half.

Your integral represents the volume of the solid whose base is your region R. The symmetry allows you to find the volume over the upper half ([0, 1] X [0, 1]) and double it.

 

[brmath]

The only obvious symmetries to me are that the x reflects across the x-axis and the $y^2$ across the y axis. Using either you get $\int \int(2 + xy^2) dA = 2\int_0^1 \int_0^1 (2 + xy^2)dxdy$; or you can do the same integral dydx, but you could anyway.

Although I am a great fan of using symmetry to dodge work, I don't see how this helps you much. Perhaps there is some symmetry I don't see.

 

[Mark44]

The only obvious symmetries to me are that the x reflects across the x-axis and the $y^2$ across the y axis.

?

The solid whose volume is being found has an upper surface of z = f(x, y) = 2 + xy². Pretty clearly, f(x, -y) = f(x, y), which means there is symmetry across the x-z plane, as I said in my previous post.

Using either you get $\int \int(2 + xy^2) dA = 2\int_0^1 \int_0^1 (2 + xy^2)dxdy$; or you can do the same integral dydx, but you could anyway.

Although I am a great fan of using symmetry to dodge work, I don't see how this helps you much. Perhaps there is some symmetry I don't see.

 

[Office_Shredder]

Mark, I think the point is that the new integral to evaluate is literally just as difficult to evaluate as the original one, unless you find calculating (-1)³ to be unusually hard. I don't see any way to reduce the integral more than that though, so I guess it's what they intend you to do.

 

[Mark44]

I agree. I think it's just an exercise to get the student to use symmetry, not that it saves a lot of work. That's my sense at any rate.

 

[brmath]

I agree. I think it's just an exercise to get the student to use symmetry, not that it saves a lot of work. That's my sense at any rate.

I agree with both of you, and think this is not a useful example. There are many good examples where the symmetry would really help, so any students solving it would learn to keep an eye out for symmetries.